npleted 25 out of 36 Submit on 8 of 36 A sample of an ideal gas has a volume of 3.80 L at 12.60 °C and 1.50 atm. What is the volume of the gas at 18.80 °C and 0.987 atm? V = L

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Chapter1: Chemical Foundations
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**Ideal Gas Law Problem: Determining the Volume at Different Conditions**

**Problem Statement:**

A sample of an ideal gas has a volume of 3.80 L at 12.60°C and 1.50 atm. What is the volume of the gas at 18.80°C and 0.987 atm?

\[ V = \_\_\_\_\_\_\_\_\_ \text{L} \]

**Explanation:**
This problem deals with the properties of an ideal gas and how its volume changes when the temperature and pressure change. To solve this, we can use the combined gas law, which is formulated as:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \( P_1 \), \( V_1 \), and \( T_1 \) are the initial pressure, volume, and temperature.
- \( P_2 \), \( V_2 \), and \( T_2 \) are the final pressure, volume, and temperature.

**Given Data:**
- Initial Volume \( V_1 = 3.80 \text{ L} \)
- Initial Temperature \( T_1 = 12.60^\circ\text{C} = 12.60 + 273.15 = 285.75 \text{ K} \)
- Initial Pressure \( P_1 = 1.50 \text{ atm} \)
- Final Temperature \( T_2 = 18.80^\circ\text{C} = 18.80 + 273.15 = 291.95 \text{ K} \)
- Final Pressure \( P_2 = 0.987 \text{ atm} \)

**Objective:**
- Determine the final volume \( V_2 \).

This involves finding the volume \( V \) when the gas is at the new specified conditions of temperature and pressure using the formula provided.
Transcribed Image Text:**Ideal Gas Law Problem: Determining the Volume at Different Conditions** **Problem Statement:** A sample of an ideal gas has a volume of 3.80 L at 12.60°C and 1.50 atm. What is the volume of the gas at 18.80°C and 0.987 atm? \[ V = \_\_\_\_\_\_\_\_\_ \text{L} \] **Explanation:** This problem deals with the properties of an ideal gas and how its volume changes when the temperature and pressure change. To solve this, we can use the combined gas law, which is formulated as: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Where: - \( P_1 \), \( V_1 \), and \( T_1 \) are the initial pressure, volume, and temperature. - \( P_2 \), \( V_2 \), and \( T_2 \) are the final pressure, volume, and temperature. **Given Data:** - Initial Volume \( V_1 = 3.80 \text{ L} \) - Initial Temperature \( T_1 = 12.60^\circ\text{C} = 12.60 + 273.15 = 285.75 \text{ K} \) - Initial Pressure \( P_1 = 1.50 \text{ atm} \) - Final Temperature \( T_2 = 18.80^\circ\text{C} = 18.80 + 273.15 = 291.95 \text{ K} \) - Final Pressure \( P_2 = 0.987 \text{ atm} \) **Objective:** - Determine the final volume \( V_2 \). This involves finding the volume \( V \) when the gas is at the new specified conditions of temperature and pressure using the formula provided.
### Freezing Point Depression of Aqueous Solutions

**Objective:**
Assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points.

**Instructions:**
Drag and drop the given compounds in the box according to their freezing points, starting from the highest freezing point at the top to the lowest freezing point at the bottom.

**Compounds to Arrange:**
- \( \text{Al}_2(\text{SO}_4)_3 \)
- \( \text{K}_3\text{PO}_4 \)
- \( \text{KBr} \)
- \( \text{MgCl}_2 \)

**Answer Bank:**
The compounds are available in an answer bank at the bottom for easy selection.

**Diagram:**
There is an empty rectangular box provided for the arrangement of the compounds. The top of the box is labeled "Highest freezing point" and the bottom is labeled "Lowest freezing point."

**Concept Explanation:**

The freezing point of an aqueous solution is affected by the presence of solutes; this is referred to as freezing point depression. The degree of freezing point depression depends on the number of particles into which the solute dissociates in solution.

* Dissociation refers to how a compound separates into ions when dissolved in water.
* The more particles the solute dissociates into, the greater the freezing point depression.

**Steps to Solve:**

1. Identify the dissociation of each compound in water:
  - \( \text{Al}_2(\text{SO}_4)_3 \) dissociates into \( 2\text{Al}^{3+} \) and \( 3\text{SO}_4^{2-} \) (total of 5 ions).
  - \( \text{K}_3\text{PO}_4 \) dissociates into \( 3\text{K}^{+} \) and \( \text{PO}_4^{3-} \) (total of 4 ions).
  - \( \text{KBr} \) dissociates into \( \text{K}^{+} \) and \( \text{Br}^{-} \) (total of 2 ions).
  - \( \text{MgCl}_2 \) dissociates into \( \text{Mg}^{2+} \) and \( 2\text{Cl}^{-} \) (total of 3 ions
Transcribed Image Text:### Freezing Point Depression of Aqueous Solutions **Objective:** Assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points. **Instructions:** Drag and drop the given compounds in the box according to their freezing points, starting from the highest freezing point at the top to the lowest freezing point at the bottom. **Compounds to Arrange:** - \( \text{Al}_2(\text{SO}_4)_3 \) - \( \text{K}_3\text{PO}_4 \) - \( \text{KBr} \) - \( \text{MgCl}_2 \) **Answer Bank:** The compounds are available in an answer bank at the bottom for easy selection. **Diagram:** There is an empty rectangular box provided for the arrangement of the compounds. The top of the box is labeled "Highest freezing point" and the bottom is labeled "Lowest freezing point." **Concept Explanation:** The freezing point of an aqueous solution is affected by the presence of solutes; this is referred to as freezing point depression. The degree of freezing point depression depends on the number of particles into which the solute dissociates in solution. * Dissociation refers to how a compound separates into ions when dissolved in water. * The more particles the solute dissociates into, the greater the freezing point depression. **Steps to Solve:** 1. Identify the dissociation of each compound in water: - \( \text{Al}_2(\text{SO}_4)_3 \) dissociates into \( 2\text{Al}^{3+} \) and \( 3\text{SO}_4^{2-} \) (total of 5 ions). - \( \text{K}_3\text{PO}_4 \) dissociates into \( 3\text{K}^{+} \) and \( \text{PO}_4^{3-} \) (total of 4 ions). - \( \text{KBr} \) dissociates into \( \text{K}^{+} \) and \( \text{Br}^{-} \) (total of 2 ions). - \( \text{MgCl}_2 \) dissociates into \( \text{Mg}^{2+} \) and \( 2\text{Cl}^{-} \) (total of 3 ions
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